Problem: Simplify the following expression: $y = \dfrac{-5x^2- 1x+4}{-5x + 4}$
Answer: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-5)}{(4)} &=& -20 \\ {a} + {b} &=& &=& {-1} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-20$ and add them together. Remember, since $-20$ is negative, one of the factors must be negative. The factors that add up to ${-1}$ will be your ${a}$ and ${b}$ When ${a}$ is ${4}$ and ${b}$ is ${-5}$ $ \begin{eqnarray} {ab} &=& ({4})({-5}) &=& -20 \\ {a} + {b} &=& {4} + {-5} &=& -1 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-5}x^2 +{4}x) + ({-5}x +{4}) $ Factor out the common factors: $ x(-5x + 4) + 1(-5x + 4)$ Now factor out $(-5x + 4)$ $ (-5x + 4)(x + 1)$ The original expression can therefore be written: $ \dfrac{(-5x + 4)(x + 1)}{-5x + 4}$ We are dividing by $-5x + 4$ , so $-5x + 4 \neq 0$ Therefore, $x \neq \frac{4}{5}$ This leaves us with $x + 1; x \neq \frac{4}{5}$.